Born-Haber Cycles

This lesson covers Born-Haber cycles in detail, including lattice enthalpy, electron affinity, ionisation energy and atomisation energy. You will learn to construct Born-Haber cycles and use them to calculate unknown enthalpy values. This is a key topic in AQA specification 3.1.4.

---

What Is a Born-Haber Cycle?

A Born-Haber cycle is a special enthalpy cycle (application of Hess's law) used to calculate the lattice enthalpy of an ionic compound. It breaks the formation of an ionic compound from its elements into a series of steps, each with a known enthalpy change.

---

Key Enthalpy Terms

Lattice Enthalpy of Formation (ΔlattH°)

The enthalpy change when one mole of an ionic lattice is formed from its gaseous ions under standard conditions.

Na⁺(g) + Cl⁻(g) → NaCl(s) ΔlattH° = −787 kJ mol⁻¹

This is always exothermic (negative) because forming the lattice releases energy.

Note: Some textbooks define lattice enthalpy as the energy to break the lattice (lattice dissociation enthalpy), which is the reverse process and always endothermic (+787 kJ mol⁻¹ for NaCl). AQA uses the lattice formation convention but you should be comfortable with both.

First Ionisation Energy (IE₁)

The energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.

Na(g) → Na⁺(g) + e⁻ IE₁ = +496 kJ mol⁻¹

Always endothermic (positive).

Second Ionisation Energy (IE₂)

The energy required to remove a second electron from one mole of gaseous 1+ ions.

Mg⁺(g) → Mg²⁺(g) + e⁻ IE₂ = +1451 kJ mol⁻¹

First Electron Affinity (EA₁)

The enthalpy change when one mole of gaseous atoms each gains one electron to form one mole of gaseous 1− ions.

Cl(g) + e⁻ → Cl⁻(g) EA₁ = −349 kJ mol⁻¹

First electron affinity is usually exothermic (negative) because the incoming electron is attracted to the nucleus.

Second Electron Affinity (EA₂)

The enthalpy change when one mole of gaseous 1− ions each gains a further electron to form 2− ions.

O⁻(g) + e⁻ → O²⁻(g) EA₂ = +798 kJ mol⁻¹

Second electron affinity is always endothermic (positive) because the incoming electron is being added to an already negative ion, which repels it.

Enthalpy of Atomisation (ΔatH°)

As defined in Lesson 7:

• Na(s) → Na(g) ΔatH° = +107 kJ mol⁻¹
• ½Cl₂(g) → Cl(g) ΔatH° = +121 kJ mol⁻¹

Enthalpy of Formation (ΔfH°)

• Na(s) + ½Cl₂(g) → NaCl(s) ΔfH° = −411 kJ mol⁻¹

---

Constructing a Born-Haber Cycle for NaCl

The cycle links the formation of NaCl(s) from its elements to the lattice enthalpy via intermediate steps:

                    Na⁺(g) + Cl(g) + e⁻
                   /                      \
          IE₁(Na)                    EA₁(Cl)
           +496                       −349
                 /                        \
    Na(g) + Cl(g)                    Na⁺(g) + Cl⁻(g)
         |                                    |
   ΔatH°(Cl)                          ΔlattH°(NaCl)
      +121                               ???
         |                                    |
    Na(g) + ½Cl₂(g)                          |
         |                                    |
   ΔatH°(Na)                                 |
      +107                                   |
         |                                    |
    Na(s) + ½Cl₂(g) ————ΔfH°————→     NaCl(s)
                      −411

Applying Hess's Law

Going from Na(s) + ½Cl₂(g) to NaCl(s):

Route 1 (direct): ΔfH° = −411 kJ mol⁻¹

Route 2 (via gaseous ions): ΔatH°(Na) + ΔatH°(Cl) + IE₁(Na) + EA₁(Cl) + ΔlattH°

By Hess's law: Route 1 = Route 2

−411 = (+107) + (+121) + (+496) + (−349) + ΔlattH°

−411 = +375 + ΔlattH°

ΔlattH° = −411 − 375 = −786 kJ mol⁻¹

(Literature value: −787 kJ mol⁻¹ — the small difference is due to rounding.)

Exam Tip: When constructing Born-Haber cycles, work systematically. List all the steps, assign the correct sign to each enthalpy value, and use Hess's law to find the unknown. Always check that your signs are consistent.

---

Worked Example: Born-Haber Cycle for MgO

Question: Construct a Born-Haber cycle for MgO and calculate the lattice enthalpy.

Given data:

Step	Enthalpy Change (kJ mol⁻¹)
ΔatH°(Mg)	+148
IE₁(Mg)	+738
IE₂(Mg)	+1451
ΔatH°(O) = ½ × bond enthalpy of O₂	+249
EA₁(O)	−141
EA₂(O)	+798
ΔfH°(MgO)	−601

The steps from elements to ionic lattice: